# Maximum Distance? 1d Kinematics

1. Oct 11, 2006

### Kildars

A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.5 m/s. The car is a distance d away. The bear is 24 m behind the tourist and running at 6.0 m/s. The tourist reaches the car safely. What is the maximum possible value for d?

In meters.

I know the bear is at D+24.. I don't really know how to start this though.

2. Oct 11, 2006

### kreil

Set up equations for the positions at any time t for the tourist and bear:

$$x_{tourist}(t)=d-4.5t$$

$$x_{bear}(t)=(d+24)-6.0t$$

Now if you equate the two equations, you can find how long it takes for the bear to catch the tourist (when their positions are equal):

$$d-4.5t=d+24-6t \implies 1.5t=24 \implies t=16s$$

So the most time the tourist can run before getting caught is 16s, and since he is moving at 4.5m/s he goes a distance of 72m before getting caught. So the maximum of d is 72, or else the tourist will get caught.

Cheers,
Josh

3. Oct 11, 2006

### Kildars

Thank you Josh, I understand now.

4. Oct 11, 2006

### quasar987

The "extreme case" is when the bear and the tourist reach the car at the same time. This happens when the bear reaches the car in the same time that the tourist reaches the car. The toursist takes a time $\Delta t = D/4.5$ to reach the car. So you want to use the equation of kinematics

$$x(t)=x_0+v_0\Delta t$$

for the motion of the bear in the case where the bear reaches the car in a time $\Delta t = D/4.5$, and solve for D.